Answer: E(Y) = 1.6 and Var(Y)=1.12
Step-by-step explanation:
Since we have given that
X 0 1 2
P(X) 0.4 0.4 0.2
Here, number of games = 2
So, [tex]Y=X_1+X_2[/tex]
Since [tex]X_1\ and\ X_2[/tex] are independent variables.
so, [tex]E[Y]=2E[X]\\\\Var[Y]=2Var[X][/tex]
So, we get that
[tex]E(X)=0.4\times 0+0.4\times 1+0.2\times 2=0.8\\\\and Var[x]=E[x^2]-(E[x])^2\\\\E[x^2]=0\times 0.4+1\times 0.4+4\times 0.2=1.2\\\\So, Var[x]=1.2-(0.8)^2\\\\Var[x]=1.2-0.64=0.56[/tex]
So, E[y]=2×0.8=1.6
and Var[y]=2×0.56=1.12
Hence, E(Y) = 1.6 and Var(Y)=1.12